Solve x^2 + 6x + 4=0?
1 answer:
You need to use the quadratic formula for this question.
In your equation, a is 1, b is 6, and c is 4. (I couldn't find a plus minus sign, so that's what the +- means, sorry if there's confusion.)
Sub those into the quadratic formula, and
For the first answer, we should get
x1=-0.7639, when you add the √6^2-4(1)(4)
For the second answer, we get
x2=-5.2361 when we subtract the √6^2-4(1)(4)
Therefore, the x values that make this equation equal to 0 should be approximately -0.7639 and -5.2361.
The more specific answers are
x1= -3+√5 and x2=-3-√5.
In your equation, a is 1, b is 6, and c is 4. (I couldn't find a plus minus sign, so that's what the +- means, sorry if there's confusion.)
Sub those into the quadratic formula, and
For the first answer, we should get
x1=-0.7639, when you add the √6^2-4(1)(4)
For the second answer, we get
x2=-5.2361 when we subtract the √6^2-4(1)(4)
Therefore, the x values that make this equation equal to 0 should be approximately -0.7639 and -5.2361.
The more specific answers are
x1= -3+√5 and x2=-3-√5.
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Answer:
Step-by-step explanation:
The given rectangle has diagonals have the endpoints P(-3, -2) ,I(4, -7) and A(4, -2) ,D(-3, -7)
The diagonals of the rectangle bisect each other so we use the midpoint formula to find their point of intersection.
The midpoint formula is;
We use any pair of endpoints of the diagonals to find the point of intersection.
Using A(4, -2) ,D(-3, -7)
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